TITLE: Easiest proof for showing finite etale (analytic) quotients of algebraic varieties are algebraic
QUESTION [8 upvotes]: Let $X$ be an algebraic variety over $\mathbb C$. Let $X^{an}\to Y$ be a finite etale morphism with $Y$ a complex analytic space.
I read somewhere that $Y$ algebraizes, ie, $Y=V^{an}$ for some algebraic space $V$ over $\mathbb C$.  (Edit: I previously wrote $Y$ algebraic variety. For my purposes, I just want $Y$ to be an algebraic space.)
Why is this, and what is an "easy" proof for this? (Consequently, the morphism $X^{an}\to Y = V^{an}$ also algebraizes by Riemann's existence theorem. Actually, this is not right. It algebraizes in the sense that it comes from an algebraic morphism $W\to V$, but $W$ might not be isomorphic to $X$. )

REPLY [6 votes]: Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question 

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8. 

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding  all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).